Problem: Find the domain of the function
\[f(x) = \sqrt{1 - \sqrt{2 - \sqrt{3 - x}}}.\]
Solution: The function $f(x) = \sqrt{1 - \sqrt{2 - \sqrt{3 - x}}}$ is defined only when
\[1 - \sqrt{2 - \sqrt{3 - x}} \ge 0,\]or
\[\sqrt{2 - \sqrt{3 - x}} \le 1. \quad (*)\]Squaring both sides, we get
\[2 - \sqrt{3 - x} \le 1.\]Then
\[\sqrt{3 - x} \ge 1.\]Squaring both sides, we get
\[3 - x \ge 1,\]so $x \le 2.$

Also, for $(*)$ to hold, we must also have
\[2 - \sqrt{3 - x} \ge 0.\]Then $\sqrt{3 - x} \le 2.$  Squaring both sides, we get
\[3 - x \le 4,\]so $x \ge -1.$

Hence, the domain of $f(x)$ is $\boxed{[-1,2]}.$